Note: scroll to bottom to find answers to examples with an astrix
Stoichiometry is just like a baking recipe. (ex:If a recipe calls for 2 eggs and 3 cups of flour, but you only have 1 egg, then you can only use 1.5 cups of flour to make the reicpe proportional). So, if the reacion (below) requires 8 moles of NaCl, then 8 moles of Na and 4 moles of Cl2 are required to make the reaction possible. .
Equation Given: 2Na + Cl2 --> 2NaCl
-multiply all elements by 4 (to get 8 moles of NaCl [you must multiply all, never just one]).
Final Equation: 8Na + 4Cl2 --> 8NaCl
Beginning Steps to Solving a Stoichiometry Problem:
1) Write Equation
2) Balance Equation
3) Find Out How Many Moles are Necessary
4) Use Molar Mass to Convert to Grams
Finished Example: Equation Given: Fe + O2 --> Fe2O3 Balance: 4Fe + 3O2 --> 2Fe2O3 How Many Grams of Fe + O2 are Required to Make 10 Moles of Fe2O3?: Use Molar Mass to Convret to Grams:
20 mol Fe x (55.845g/1mol Fe) = 1116.9 grams
15 mol O2 x (31.998g/1mol O2) = 479.9 grams
Complete Example By Using Beginning Steps:*
How many grams of H2 are required to make 75 grams of NH3 in the following reaction?
N2 + H2 --> NH3
Alternative Way to Complete Example (combine steps):
75g NH3 (1mol NH3/17.036g) x (3mol H2/2mol NH3) x (2.02g H2/1mol H20) = 13.3 grams of H2
REMEMBER: make sure all measurements balance out (i.e. grams on top and on bottom, etc.)
Limiting Reactant Problem:
If you have 25.2 grams of N2 + 15.2 grams H2, is your limiting reactant in the production of NH3?
25.2 of N2 (1 mol/28g) ( 2 mol/1 mol) (17.03g/1 mol) = 30.654 g
15.2 (1mol/2.02g) (2 mol/3 mol) (17.03/1ol) = 85.431g Limiting Reactant = 30.654g
Mole to Volume Conversion
1 mole
22.4 liters
To find the volume of a reactanct or product, just add this conversion to the end of the problem to make the final answer in liters.
Steps to Find Percentage Composition:
1) Find Individual Amount of Grams Per Element
2) Find Total Number of Grams
3) Create Proportions to Find Percentages
Finished Example: Given: Ca3(PO4)2 Individual Amount of Grams: Ca3 = 120 (40 x 3) P = 62 (31 x 2) O4 = 128 (16 x 8) Find Total Number of Grams: 310 grams/mole Create Proportions: Ca3 = 38% (120/310) P= 26% (62/310) O4 = 42% (128/310)
Steps to Find Emperical Formula:
1) Find the Number of Moles of Each Compound
2) Divide by the Smallest Number of Moles to Get a Ratio
3) Apply that Ratio to the Compound
Empirical Formula - smallest whole number taio of atoms in a compound
Finished Example:
A compound is found to be 34.8 O and 65.2 g As by mass. Find the Emp. Form.
Steps to Fine Molecular Formula:
1) Find the Emp. Form.
2) Find the Mass of the Emp. Form.
3) Divide the Mass of the Molecular Mass by the Mass of the Emp. Form.
4) Multiply Your Emp. Form. by the Answer From Step 3
Molecular Formula - actual formula of a compound
Finished Example: Find Empirical Formula: CH Find Mass of Empirical Formula: 13 grams/mole Divide: 78/13 = 6 Multiply by Step 3: C6H6
Answers to Example Problems:
75g NH3 x (1mol NH3 / 17.036g) x (3mol H2 / 2mol NH3) x (2.02g H2 / 1mol H2) = 13.3 grams of H2
Chapter 9 - Stoichiometry
Stoichiometry is just like a baking recipe. (ex:If a recipe calls for 2 eggs and 3 cups of flour, but you only have 1 egg, then you can only use 1.5 cups of flour to make the reicpe proportional). So, if the reacion (below) requires 8 moles of NaCl, then 8 moles of Na and 4 moles of Cl2 are required to make the reaction possible. .
Equation Given: 2Na + Cl2 --> 2NaCl-multiply all elements by 4 (to get 8 moles of NaCl [you must multiply all, never just one]).
Final Equation: 8Na + 4Cl2 --> 8NaCl
1) Write Equation
2) Balance Equation
3) Find Out How Many Moles are Necessary
4) Use Molar Mass to Convert to Grams
Equation Given: Fe + O2 --> Fe2O3
Balance: 4Fe + 3O2 --> 2Fe2O3
How Many Grams of Fe + O2 are Required to Make 10 Moles of Fe2O3?:
Use Molar Mass to Convret to Grams:
20 mol Fe x (55.845g/1mol Fe) = 1116.9 grams
15 mol O2 x (31.998g/1mol O2) = 479.9 grams
Complete Example By Using Beginning Steps:*
How many grams of H2 are required to make 75 grams of NH3 in the following reaction?
N2 + H2 --> NH3
Alternative Way to Complete Example (combine steps):
75g NH3 (1mol NH3/17.036g) x (3mol H2/2mol NH3) x (2.02g H2/1mol H20) = 13.3 grams of H2
REMEMBER: make sure all measurements balance out (i.e. grams on top and on bottom, etc.)
Limiting Reactant Problem:
If you have 25.2 grams of N2 + 15.2 grams H2, is your limiting reactant in the production of NH3?
25.2 of N2 (1 mol/28g) ( 2 mol/1 mol) (17.03g/1 mol) = 30.654 g
15.2 (1mol/2.02g) (2 mol/3 mol) (17.03/1ol) = 85.431g
Limiting Reactant = 30.654g
1) Find Individual Amount of Grams Per Element
2) Find Total Number of Grams
3) Create Proportions to Find Percentages
Given: Ca3(PO4)2
Individual Amount of Grams: Ca3 = 120 (40 x 3) P = 62 (31 x 2) O4 = 128 (16 x 8)
Find Total Number of Grams: 310 grams/mole
Create Proportions: Ca3 = 38% (120/310) P= 26% (62/310) O4 = 42% (128/310)
1) Find the Number of Moles of Each Compound
2) Divide by the Smallest Number of Moles to Get a Ratio
3) Apply that Ratio to the Compound
Finished Example:
A compound is found to be 34.8 O and 65.2 g As by mass. Find the Emp. Form.
34.8g O (1 mol/ 16 g) = 2.17 mol
65.2g As (1 mol/75 g) = .87 mol
compound: As2O5
ratio: 1:2.5
1) Find the Emp. Form.
2) Find the Mass of the Emp. Form.
3) Divide the Mass of the Molecular Mass by the Mass of the Emp. Form.
4) Multiply Your Emp. Form. by the Answer From Step 3
Finished Example:
Find Empirical Formula: CH
Find Mass of Empirical Formula: 13 grams/mole
Divide: 78/13 = 6
Multiply by Step 3: C6H6
Answers to Example Problems:
Works Cited:
"Stoichiometry" 19 May 2009 Chem4kids //http://www.chem4kids.com/files/react_stoichio.html//